2. In a random sample of 130 World Campus students, 92 were employed full-time.

We want use these data to construct a 95% confidence interval to estimate the proportion of all World Campus students who are employed full-time. Is it appropriate to use the normal approximation method? Show how you checked assumptions.

Step-by-step explanation:

Given that sample size is 130 >30. Also by central limit theorem, we know that mean (here proportion) of all means of different samples would tend to become normal with mean = average of all means (here proportions)

Hence we can assume normality assumptions here.

Proportion sample given = 92/130 = 0.7077

The mean proportion of different samples for large sample size will follow normal with mean = sample proportion and std error = square root of p (1-p) / n

Hence mean proportion p = 0.7077

q = 1-p = 0.2923

Std error = 0.0399

For 95% confidence interval we find that z critical for 95% two tailed is 1,.96

Hence margin of error = + or - 1.96 (std error)

= 0.0782

Confidence interval = (p-margin of error, p+margin of error)

= (0.7077-0.0782,0.7077+0.0782)

= (0.6295, 0.7859)

We are 95% confident that average of sample proportions of different samples would lie within these values in the interval for large sample sizes.